Can we obtain the same result as Deutsch algorithm using a XOR gate?

Question

If I understand correctly, Deutsch's algorithm is useful to know if a function $f(x)$ is constant ($f(0)=f(1)$) or balanced ($f(0) \neq f(1)$). And the advantage is that it only requires the measurement of one quibit.

But can't we obtain the same result by using classical bits by measuring the output of a XOR gate with inputs $f(0)$ and $f(1)$?

Answer 1

The (nominal) advantage of Deutsch’s algorithm isn’t that you only need to measure one qubit. It’s that you only need to evaluate the function $f$ (which might be slow to do) once, whereas in the classical version you need to evaluate it twice before plugging the outputs into the XOR.